is over all spacetime.
So the path integral tells us to integrate over all
classical
field configurations
Φ
. Note that
Φ
does not just consist of the oneparticle states, it can have 2particle states, etc.
We can
remember this by drawing pictures for the paths – including disconnected bubbles – as we would
using Feynman rules.
Actually, we really sum over all kinds of discontinuous, disconnected
random fluctuations. In perturbation theory, only paths corresponding to sums of states of fixed
particle number contribute. Nonperturbatively, for example with bound states or situations
where multiple soft photons are relevant, particle number may not be a useful concept. The
path integral allows us to perform calculations in nonperturbative regimes.
2.4
Classical limit
As a first check on the path integral, we can take the classical limit. To do that, we need to put
back
planckover2pi1
which can be done by dimensional analysis. Since
planckover2pi1
has dimensions of action, it appears
as
(
0;
t
f

0;
t
i
)
=
N
integraldisplay
D
Φ(
x,t
)
e
i
planckover2pi1
S
[Φ]
(34)
2.
If
V
depended on
π
ˆ
and
φ
ˆ
, there might be an ordering ambiguity; this is no different than in the nonrela
tivistic case and it is conventional to define the Hamiltonian to be
Weylordered
with the
π
ˆ
operators all to the
left of the
φ
ˆ
operators.
6
Section 2
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Using the method of stationary phase we see that in the limit
planckover2pi1
→
0
, this integral is dominated
by the value of
Φ
for which
S
[Φ]
has an extremum. But
δS
= 0
is precisely the condition which
determines the EulerLagrange equations which a classical field satisfies. Therefore the only con
figuration which contributes in the classical limit is the classical solution to the equations of
motion.
In case you are not familiar with
method
of
stationary
phase
(also known as the
method of steepest descent
), it is easy to understand. The quickest way is to start with the
same integral without the
i
:
integraldisplay
D
Φ(
x,t
)
e
−
1
planckover2pi1
S
[Φ]
(35)
In this case, the integral would clearly be dominated by the
Φ
0
where
S
[Φ]
has a minimum;
everything else would give a bigger
S
[Φ]
and be infinitely more suppressed as
planckover2pi1
→
0
.
Now, when
we put the
i
back in, the same thing happens, not because the nonminimal terms are zero, but
because away from the minimum you have to sum over phases swirling around infinitely fast.
When you sum infinitely swirling phases, you also get something which goes to zero when com
pared to something with a constant phase. Another way to see it is to use the more intuitive
case with
e
−
1
planckover2pi1
S
[Φ]
. Since we expect the answer to be welldefined, it should be an analytic func
tion of
Φ
0
. So we can take
planckover2pi1
→
0
in the imaginary direction, showing that the integral is still
dominated by
S
[Φ
0
]
.